Model paper

Dean's Office Official Model Question Paper

ELE 229 · Statistical Methods and Data Modeling

Programme
BBA-F
Academic year
Semester 7
Paper type
Official Model Question
Sitting
Dean's Office Blueprint
Full marks
60
Duration
180 minutes

Tribhuvan University

Faculty of Management

Office of the Dean

Official Model Question Paper / Dean's Office Blueprint

Course: ELE 229 · Statistical Methods and Data Modeling

Level: Bachelor of Business Administration in Finance (BBA-F) · Semester 7

Full Marks: 60

Time: 3 hrs.

Candidates are required to give their answers in their own words as far as practicable. Figures in the margin indicate full marks.

Group A

Brief Answer Questions. Attempt ALL questions. (5 × 2 = 10)

[5*2=10]
  1. Define Type I error and Type II error in hypothesis testing.

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    Type I and Type II Errors

    • Type I Error (α\alpha): Rejecting the null hypothesis (H0H_0) when it is actually true (false positive). The probability of committing a Type I error is the significance level α\alpha.
    • Type II Error (β\beta): Failing to reject the null hypothesis (H0H_0) when it is actually false (false negative). The probability of correctly rejecting a false null hypothesis is statistical power (1β1 - \beta).
  2. Distinguish between Laspeyre’s and Paasche’s Price Index numbers.

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    Laspeyre’s vs. Paasche’s Price Index

    • Laspeyre’s Index (P01LP_{01}^L): Uses base-year quantities (q0q_0) as weights: P01L=p1q0p0q0×100P_{01}^L = \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100. It tends to overestimate price increases (upward bias) because it ignores consumer substitution toward cheaper goods.
    • Paasche’s Index (P01PP_{01}^P): Uses current-year quantities (q1q_1) as weights: P01P=p1q1p0q1×100P_{01}^P = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100. It tends to underestimate price increases (downward bias) by over-weighting goods whose relative prices have fallen.
  3. State the key Gauss-Markov assumptions of the Classical Linear Regression Model (CLRM).

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    Gauss-Markov Assumptions of CLRM

    1. Linearity in Parameters: The model is linear in coefficients: Y=β0+β1X+ϵY = \beta_0 + \beta_1 X + \epsilon.
    2. Zero Conditional Mean: E(ϵX)=0E(\epsilon \mid X) = 0 (error disturbances have expected mean zero).
    3. Homoskedasticity: Var(ϵiX)=σ2\text{Var}(\epsilon_i \mid X) = \sigma^2 (constant error variance across all observations).
    4. No Autocorrelation: Cov(ϵi,ϵj)=0\text{Cov}(\epsilon_i, \epsilon_j) = 0 for iji \ne j (uncorrelated error terms).
    5. Exogeneity: Cov(Xi,ϵi)=0\text{Cov}(X_i, \epsilon_i) = 0 (independent variables are uncorrelated with error terms).

    When these hold, OLS estimators are BLUE (Best Linear Unbiased Estimators).

  4. What is the coefficient of determination (R2R^2) and why is the adjusted coefficient of determination (Rˉ2\bar{R}^2) necessary?

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    R2R^2 and Adjusted Rˉ2\bar{R}^2

    • R2R^2 (Coefficient of Determination): Measures the proportion of total variation in the dependent variable explained by the regression model: R2=SSRSST=1SSESSTR^2 = \frac{SSR}{SST} = 1 - \frac{SSE}{SST}.
    • Need for Rˉ2\bar{R}^2: In multiple regression, R2R^2 mechanically increases whenever any additional independent variable is added, even if irrelevant. Adjusted Rˉ2\bar{R}^2 penalizes the addition of non-significant variables by adjusting for degrees of freedom: Rˉ2=1(1R2)n1nk1\bar{R}^2 = 1 - \left(1 - R^2\right) \frac{n - 1}{n - k - 1}.
  5. Explain the five-number summary displayed in a Box and Whisker plot.

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    Five-Number Summary of Box and Whisker Plot

    A Box and Whisker plot visualizes distribution, central tendency, and dispersion using:

    1. Minimum Value: Lowest data point within the lower fence (Q11.5×IQRQ_1 - 1.5 \times IQR).
    2. First Quartile (Q1Q_1): 25th percentile (bottom of the box).
    3. Median (Q2Q_2): 50th percentile (central line within the box).
    4. Third Quartile (Q3Q_3): 75th percentile (top of the box).
    5. Maximum Value: Highest data point within the upper fence (Q3+1.5×IQRQ_3 + 1.5 \times IQR). Data points outside the fences are identified as outliers.

Group B

Short Answer Questions. Attempt any THREE questions. (3 × 10 = 30)

[3*10=30]
  1. A commercial bank tests whether the average loan processing turnaround time differs between two regional hubs: Pokhara Hub and Biratnagar Hub. Random samples of SME credit applications produce the following statistics:

    • Pokhara Hub: n1 = 12, mean1 = 14.5 hours, s1 = 2.4 hours
    • Biratnagar Hub: n2 = 10, mean2 = 17.2 hours, s2 = 2.8 hours Assuming normal distributions and equal population variances, test at the 5% significance level (alpha = 0.05) whether there is a statistically significant difference in mean processing turnaround times. (Given: critical t(0.025, 20) = 2.086).
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    Numerical Problem: Independent Two-Sample t-Test

    1. Hypotheses Formulation

    • Null Hypothesis (H0H_0): μ1μ2=0\mu_1 - \mu_2 = 0 (There is no significant difference in mean processing turnaround times between Pokhara and Biratnagar hubs).
    • Alternative Hypothesis (H1H_1): μ1μ20\mu_1 - \mu_2 \ne 0 (There is a statistically significant difference; two-tailed test).
    • Significance level: α=0.05\alpha = 0.05.
    • Degrees of freedom: df=n1+n22=12+102=20df = n_1 + n_2 - 2 = 12 + 10 - 2 = 20.

    2. Calculation of Pooled Variance (sp2s_p^2)

    sp2=(n11)s12+(n21)s22n1+n22s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}
    s12=(2.4)2=5.76s_1^2 = (2.4)^2 = 5.76
    s22=(2.8)2=7.84s_2^2 = (2.8)^2 = 7.84
    sp2=(121)(5.76)+(101)(7.84)20=11(5.76)+9(7.84)20=63.36+70.5620=133.9220=6.696s_p^2 = \frac{(12 - 1)(5.76) + (10 - 1)(7.84)}{20} = \frac{11(5.76) + 9(7.84)}{20} = \frac{63.36 + 70.56}{20} = \frac{133.92}{20} = 6.696
    sp=6.6962.5877 hourss_p = \sqrt{6.696} \approx 2.5877\text{ hours}

    3. Standard Error of the Difference (SESE)

    SE=sp1n1+1n2=2.5877×112+110=2.5877×0.08333+0.1000SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 2.5877 \times \sqrt{\frac{1}{12} + \frac{1}{10}} = 2.5877 \times \sqrt{0.08333 + 0.1000}
    SE=2.5877×0.18333=2.5877×0.428171.108 hoursSE = 2.5877 \times \sqrt{0.18333} = 2.5877 \times 0.42817 \approx 1.108\text{ hours}

    4. Calculated t-Statistic (tcalt_{cal})

    tcal=xˉ1xˉ2SE=14.517.21.108=2.71.1082.437t_{cal} = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{14.5 - 17.2}{1.108} = \frac{-2.7}{1.108} \approx \mathbf{-2.437}
    tcal=2.437|t_{cal}| = 2.437

    5. Decision & Managerial Conclusion

    • Critical Value: For df=20df = 20 at α=0.05\alpha = 0.05 (two-tailed), tcrit=2.086t_{crit} = 2.086.
    • Decision Rule: Reject H0H_0 if tcal>tcrit|t_{cal}| > t_{crit}.
    • Since tcal=2.437>2.086|t_{cal}| = 2.437 > 2.086, we reject H0H_0 at the 5% significance level (p<0.05p < 0.05).
    • Conclusion: There is a statistically significant difference in SME loan turnaround times between Pokhara and Biratnagar hubs. Pokhara operates significantly faster (14.5 hours vs. 17.2 hours). Bank management should investigate procedural bottlenecks and underwriting delays in Biratnagar to standardize processing velocity.
  2. A commercial bank surveyed 300 retail banking customers across three age brackets to evaluate channel preference (Mobile Banking App vs. Physical Branch):

    Age Bracket Mobile App Physical Branch Total
    Young (<30 yrs) 90 30 120
    Middle-Aged (30-50 yrs) 65 45 110
    Senior (>50 yrs) 25 45 70
    Total 180 120 300

    Test at the 5% significance level (alpha = 0.05) whether customer channel preference is independent of age group. (Given critical Chi-Square(0.05, 2) = 5.991).

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    Numerical Problem: Chi-Square Test of Independence

    1. Hypotheses Formulation

    • Null Hypothesis (H0H_0): Customer channel preference (Mobile App vs. Physical Branch) is independent of age group.
    • Alternative Hypothesis (H1H_1): Customer channel preference is dependent on age group.
    • Significance level: α=0.05\alpha = 0.05.
    • Degrees of freedom: df=(r1)(c1)=(31)(21)=2×1=2df = (r - 1)(c - 1) = (3 - 1)(2 - 1) = 2 \times 1 = 2.

    2. Calculation of Expected Frequencies (Eij=Ri×CjNE_{ij} = \frac{R_i \times C_j}{N})

    • Row 1 (Young, R1=120R_1 = 120):
      • E11(Mobile)=120×180300=72.0E_{11} (\text{Mobile}) = \frac{120 \times 180}{300} = 72.0
      • E12(Branch)=120×120300=48.0E_{12} (\text{Branch}) = \frac{120 \times 120}{300} = 48.0
    • Row 2 (Middle-Aged, R2=110R_2 = 110):
      • E21(Mobile)=110×180300=66.0E_{21} (\text{Mobile}) = \frac{110 \times 180}{300} = 66.0
      • E22(Branch)=110×120300=44.0E_{22} (\text{Branch}) = \frac{110 \times 120}{300} = 44.0
    • Row 3 (Senior, R3=70R_3 = 70):
      • E31(Mobile)=70×180300=42.0E_{31} (\text{Mobile}) = \frac{70 \times 180}{300} = 42.0
      • E32(Branch)=70×120300=28.0E_{32} (\text{Branch}) = \frac{70 \times 120}{300} = 28.0

    3. Chi-Square Test Statistic Computation (χ2=(OE)2E\chi^2 = \sum \frac{(O - E)^2}{E})

    Cell (Age, Channel) Observed (OO) Expected (EE) OEO - E (OE)2(O - E)^2 (OE)2E\frac{(O - E)^2}{E}
    Young, Mobile 90 72.0 +18.0 324.0 4.500
    Young, Branch 30 48.0 -18.0 324.0 6.750
    Middle, Mobile 65 66.0 -1.0 1.0 0.015
    Middle, Branch 45 44.0 +1.0 1.0 0.023
    Senior, Mobile 25 42.0 -17.0 289.0 6.881
    Senior, Branch 45 28.0 +17.0 289.0 10.321
    Total 300 300.0 0.0 28.490
    χcal2=28.490\chi^2_{cal} = \mathbf{28.490}

    4. Decision & Strategic Implications

    • Critical Value: For df=2df = 2 at α=0.05\alpha = 0.05, χcrit2=5.991\chi^2_{crit} = 5.991.
    • Decision: Since χcal2=28.490>5.991\chi^2_{cal} = 28.490 > 5.991, we reject H0H_0 at the 5% significance level (p<0.001p < 0.001).
    • Conclusion: Banking channel preference is highly dependent on age. Younger clients disproportionately adopt digital mobile banking (75%75\% vs. expected 60%60\%), whereas senior citizens heavily rely on physical branch interactions (64.3%64.3\% vs. expected 40%40\%). The bank should maintain hybrid branch services while designing targeted digital onboarding and accessibility features for seniors.
  3. Examine econometric diagnostic procedures for multiple regression analysis: analyze the causes, consequences, detection methods, and remedial treatments for Multicollinearity and Heteroskedasticity.

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    Econometric Diagnostics: Multicollinearity and Heteroskedasticity

    In multiple linear regression Y=β0+β1X1+β2X2++βkXk+ϵY = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \dots + \beta_k X_k + \epsilon, departures from CLRM assumptions undermine statistical inference:

    +-----------------------+------------------------+-------------------------------+
    | Diagnostic Dimension  | Multicollinearity      | Heteroskedasticity            |
    +-----------------------+------------------------+-------------------------------+
    | Violated Assumption   | Regressors independent | Error variance constant       |
    | Nature of Violation   | High linear correlation| Non-constant variance Var(e|X)|
    | OLS Estimator Quality | Unbiased but large SE  | Unbiased but inefficient      |
    | Hypothesis Tests      | Artificially low t-stat| Biased SE; invalid t/F tests  |
    +-----------------------+------------------------+-------------------------------+
    

    1. Multicollinearity

    • Causes: Occurs when two or more explanatory variables share strong linear relationships (e.g., including both household disposable income and wealth in a loan demand model).
    • Consequences:
      • OLS coefficients remain unbiased and consistent, but standard errors become inflated.
      • Individual t-statistics are insignificant despite a very high overall R2R^2 and highly significant F-statistic.
      • Estimates are hypersensitive to small data sample alterations.
    • Detection Methods:
      • Correlation matrix inspection (r>0.80r > 0.80).
      • Variance Inflation Factor (VIF): VIFj=11Rj2VIF_j = \frac{1}{1 - R_j^2}. A rule of thumb is VIF>5VIF > 5 to 1010 signifies problematic collinearity.
    • Remedies: Drop redundant collinear variables; transform variables using first differences or ratios; employ Principal Component Analysis (PCA) or Ridge regression.

    2. Heteroskedasticity

    • Causes: Occurs when error variance varies across observations: Var(ϵi)=σi2\text{Var}(\epsilon_i) = \sigma_i^2 (common in cross-sectional firm-level banking data where large banks exhibit greater variance than small microfinance institutions).
    • Consequences:
      • OLS estimates remain unbiased, but they are no longer Best (efficient).
      • Standard errors are biased, rendering usual confidence intervals and t/F tests invalid.
    • Detection Methods:
      • Residual scatter plots (ei2e_i^2 against Y^i\hat{Y}_i).
      • Formal statistical tests: Breusch-Pagan-Godfrey test and White’s General Heteroskedasticity test.
    • Remedies:
      • Logarithmic transformations of non-linear financial variables (e.g., ln(Assets)\ln(\text{Assets})).
      • Estimation using White’s Heteroskedasticity-Consistent Robust Standard Errors.
      • Weighted Least Squares (WLS) where weights are inversely proportional to estimated error variances.
  4. (a) Define Fisher’s Ideal Index Number and prove algebraically that it satisfies both the Time Reversal Test and the Factor Reversal Test. (b) Explain the components of a financial time series and how seasonal indices are constructed.

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    (a) Fisher’s Ideal Index Number & Consistency Tests

    Fisher’s Ideal Index is the geometric mean of Laspeyre’s (LL) and Paasche’s (PP) price index numbers:

    P01F=P01L×P01P=(p1q0p0q0)×(p1q1p0q1)×100P_{01}^F = \sqrt{P_{01}^L \times P_{01}^P} = \sqrt{\left(\frac{\sum p_1 q_0}{\sum p_0 q_0}\right) \times \left(\frac{\sum p_1 q_1}{\sum p_0 q_1}\right)} \times 100

    1. Time Reversal Test (P01×P10=1P_{01} \times P_{10} = 1)

    Interchanging base year (0) and current year (1):

    P10F=(p0q1p1q1)×(p0q0p1q0)P_{10}^F = \sqrt{\left(\frac{\sum p_0 q_1}{\sum p_1 q_1}\right) \times \left(\frac{\sum p_0 q_0}{\sum p_1 q_0}\right)}
    Multiplying both price indices (excluding factor 100):
    P01F×P10F=p1q0p0q0×p1q1p0q1×p0q1p1q1×p0q0p1q0=1=1P_{01}^F \times P_{10}^F = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1} \times \frac{\sum p_0 q_1}{\sum p_1 q_1} \times \frac{\sum p_0 q_0}{\sum p_1 q_0}} = \sqrt{1} = \mathbf{1}
    Hence, Fisher’s Index satisfies the Time Reversal Test.

    2. Factor Reversal Test (P01×Q01=p1q1p0q0P_{01} \times Q_{01} = \frac{\sum p_1 q_1}{\sum p_0 q_0})

    Interchanging price (pp) and quantity (qq) to obtain the quantity index:

    Q01F=(q1p0q0p0)×(q1p1q0p1)=(p0q1p0q0)×(p1q1p1q0)Q_{01}^F = \sqrt{\left(\frac{\sum q_1 p_0}{\sum q_0 p_0}\right) \times \left(\frac{\sum q_1 p_1}{\sum q_0 p_1}\right)} = \sqrt{\left(\frac{\sum p_0 q_1}{\sum p_0 q_0}\right) \times \left(\frac{\sum p_1 q_1}{\sum p_1 q_0}\right)}
    Multiplying price and quantity indices:
    P01F×Q01F=p1q0p0q0×p1q1p0q1×p0q1p0q0×p1q1p1q0=(p1q1)2(p0q0)2=p1q1p0q0P_{01}^F \times Q_{01}^F = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1} \times \frac{\sum p_0 q_1}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_1 q_0}} = \sqrt{\frac{(\sum p_1 q_1)^2}{(\sum p_0 q_0)^2}} = \mathbf{\frac{\sum p_1 q_1}{\sum p_0 q_0}}
    Hence, Fisher’s Index satisfies the Factor Reversal Test (true value ratio).


    (b) Components of Financial Time Series

    A time series decomposes into four systematic elements under the multiplicative model Y=T×S×C×IY = T \times S \times C \times I:

    1. Secular Trend (TT): Long-term smooth progression over decades (e.g., secular growth in national banking assets).
    2. Seasonal Variation (SS): Short-term regular quarterly or monthly recurring fluctuations within a year (e.g., surge in currency in circulation before Dashain festival).
    3. Cyclical Movement (CC): Medium-term wave-like oscillations spanning 3 to 10 years tied to macroeconomic boom and recession business cycles.
    4. Irregular / Random Disturbance (II): Unpredictable shocks (e.g., natural disasters, geopolitical supply shocks).

    Seasonal Index Calculation: The Ratio-to-Moving-Average method computes a 4-quarter centered moving average to eliminate trend and cycle (T×CT \times C), isolates seasonal-irregular ratios (S×I=YT×CS \times I = \frac{Y}{T \times C}), and averages quarterly values to yield clean Seasonal Indices normalized to 400%400\%.

Group C

Comprehensive Answer / Case Analysis Question. Attempt ALL questions. (1 × 20 = 20)

[1*20=20]
  1. Comprehensive Econometric Case Analysis: Determinants of Commercial Bank Credit Growth in Nepal

    A financial econometrician investigates the quarterly loan disbursements of commercial banks in Nepal (Y, in NPR Billions) as a function of Total Customer Deposits (X1, in NPR Billions) and the Weighted Average Base Lending Rate (X2, in %). From a sample of n = 30 quarters, the estimated Ordinary Least Squares (OLS) model is:

    Y_hat = 15.40 + 0.72 X1 - 1.85 X2

    Additional regression summary statistics:

    • Standard error of beta1 = 0.09
    • Standard error of beta2 = 0.62
    • Standard error of intercept beta0 = 4.20
    • Total Sum of Squares (SST) = 4,500
    • Sum of Squared Errors (SSE) = 900
    • Regression Sum of Squares (SSR) = 3,600

    Questions: (a) Interpret the economic and statistical meaning of the regression intercept and the two partial slope coefficients (beta1 and beta2). Do the signs match theoretical economic expectations? (5 marks) (b) Calculate: (i) The coefficient of determination (R2), (ii) The adjusted coefficient of determination (R2_adj), and (iii) The standard error of the estimate (se). Interpret each metric. (5 marks) (c) Test the individual statistical significance of beta1 and beta2 at the 5% significance level (alpha = 0.05). State null and alternative hypotheses, calculate t-statistics, compare with critical t(0.025, 27) = 2.052, and draw conclusions. (5 marks) (d) Perform the overall model goodness-of-fit test using the F-test at alpha = 0.05 (Given critical F(0.05, 2, 27) = 3.35). If a commercial bank reports customer deposits of NPR 150 Billion and the prevailing base rate is 8.5%, compute the predicted loan disbursement Y_hat. (5 marks)

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    Case Analysis: Determinants of Commercial Bank Credit Growth

    (a) Interpretation of Regression Coefficients (5 Marks)

    1. Intercept (β^0=15.40\hat{\beta}_0 = 15.40):
      • When total deposits are zero (X1=0X_1 = 0) and base interest rate is zero (X2=0X_2 = 0), the baseline quarterly bank loan disbursement is estimated at NPR 15.40 Billion. Statistically, it represents autonomous credit disbursements funded via equity capital or interbank borrowings.
    2. Slope of Deposits (β^1=+0.72\hat{\beta}_1 = +0.72):
      • Holding the base rate constant, for every NPR 1.0 Billion increase in total deposits, loan disbursements increase on average by NPR 0.72 Billion (720 Million).
      • Economic Consistency: Positively signed, aligning with commercial banking theory and NRB’s Credit-Deposit (CD) prudential framework where deposit mobilization directly expands lending capacity.
    3. Slope of Base Interest Rate (β^2=1.85\hat{\beta}_2 = -1.85):
      • Holding deposit volume constant, for every 1.0 percentage point (100 bps) increase in the base lending rate, loan disbursements decrease on average by NPR 1.85 Billion.
      • Economic Consistency: Negatively signed, consistent with the classical law of demand for credit; higher borrowing costs deter corporate and consumer capital expenditures.

    (b) Goodness-of-Fit Metrics & Standard Error of Estimate (5 Marks)

    1. Coefficient of Determination (R2R^2):

    R2=SSRSST=3,6004,500=0.8000 (80.0%)R^2 = \frac{SSR}{SST} = \frac{3,600}{4,500} = \mathbf{0.8000\ (80.0\%)}

    • Interpretation: 80.0% of the total quarterly variation in commercial bank loan disbursements is explained by deposit volumes and base lending rates.

    2. Adjusted Coefficient of Determination (Rˉ2\bar{R}^2):

    Rˉ2=1(1R2)(n1nk1)\bar{R}^2 = 1 - \left(1 - R^2\right) \left(\frac{n - 1}{n - k - 1}\right)
    Where sample size n=30n = 30 and number of explanatory variables k=2k = 2 (df=nk1=3021=27df = n - k - 1 = 30 - 2 - 1 = 27):
    Rˉ2=1(10.8000)×(2927)=1(0.2000×1.07407)=10.21481=0.7852 (78.52%)\bar{R}^2 = 1 - (1 - 0.8000) \times \left(\frac{29}{27}\right) = 1 - (0.2000 \times 1.07407) = 1 - 0.21481 = \mathbf{0.7852\ (78.52\%)}

    • Interpretation: After adjusting for degrees of freedom, the model reliably accounts for 78.52% of variance, confirming high predictive power without overfitting.

    3. Standard Error of the Estimate (ses_e):

    se=SSEnk1=90027=33.3335.7735 NPR Billions_e = \sqrt{\frac{SSE}{n - k - 1}} = \sqrt{\frac{900}{27}} = \sqrt{33.333} \approx \mathbf{5.7735\text{ NPR Billion}}

    • Interpretation: The standard deviation of observed loan disbursements around the fitted regression hyperplane is approximately NPR 5.77 Billion.

    (c) Individual Hypothesis Testing (tt-Tests) at α=0.05\alpha = 0.05 (5 Marks)

    • Degrees of freedom: df=nk1=3021=27df = n - k - 1 = 30 - 2 - 1 = 27.
    • Two-tailed critical value: t0.025,27=2.052t_{0.025, 27} = 2.052.

    1. Test for Total Deposits (β1\beta_1):

    • Hypotheses: H0:β1=0H_0: \beta_1 = 0 vs. H1:β10H_1: \beta_1 \ne 0tcal,1=β^10SE(β^1)=0.720.09=+8.000t_{cal, 1} = \frac{\hat{\beta}_1 - 0}{SE(\hat{\beta}_1)} = \frac{0.72}{0.09} = \mathbf{+8.000}$
    • Decision: Since tcal,1=8.000>2.052|t_{cal, 1}| = 8.000 > 2.052, we reject H0H_0 (p<0.0001p < 0.0001). Total customer deposits exert a statistically significant positive effect on bank loan growth.

    2. Test for Base Lending Rate (β2\beta_2):

    • Hypotheses: H0:β2=0H_0: \beta_2 = 0 vs. H1:β20H_1: \beta_2 \ne 0tcal,2=β^20SE(β^2)=1.850.62=2.984t_{cal, 2} = \frac{\hat{\beta}_2 - 0}{SE(\hat{\beta}_2)} = \frac{-1.85}{0.62} = \mathbf{-2.984}$
      tcal,2=2.984|t_{cal, 2}| = 2.984
    • Decision: Since tcal,2=2.984>2.052|t_{cal, 2}| = 2.984 > 2.052, we reject H0H_0 (p<0.01p < 0.01). The base interest rate exerts a statistically significant negative effect on loan disbursements.

    (d) Overall Model Significance (FF-Test) & Loan Forecast (5 Marks)

    1. Overall FF-Test for Goodness of Fit:

    • Hypotheses: H0:β1=β2=0H_0: \beta_1 = \beta_2 = 0 (Model has no explanatory power) vs. H1:At least one βj0H_1: \text{At least one } \beta_j \ne 0.
    • Mean Square Regression (MSRMSR): MSR=SSRk=3,6002=1,800MSR = \frac{SSR}{k} = \frac{3,600}{2} = 1,800.
    • Mean Square Error (MSEMSE): MSE=SSEnk1=90027=33.333MSE = \frac{SSE}{n - k - 1} = \frac{900}{27} = 33.333.
      Fcal=MSRMSE=1,80033.333=54.00F_{cal} = \frac{MSR}{MSE} = \frac{1,800}{33.333} = \mathbf{54.00}
    • Critical Value: F0.05,(2,27)=3.35F_{0.05, (2, 27)} = 3.35.
    • Decision: Because Fcal=54.00>3.35F_{cal} = 54.00 > 3.35, we reject H0H_0 (p<0.0001p < 0.0001). The multiple regression model as a whole is exceptionally robust and statistically significant.

    2. Point Prediction for X1=150 BillionX_1 = 150\text{ Billion}, X2=8.5%X_2 = 8.5\%:

    Y^=15.40+0.72(150)1.85(8.5)\hat{Y} = 15.40 + 0.72(150) - 1.85(8.5)
    Y^=15.40+108.0015.725=107.675 NPR Billion\hat{Y} = 15.40 + 108.00 - 15.725 = \mathbf{107.675\text{ NPR Billion}}

    • Forecasting Conclusion: For a bank holding NPR 150 Billion in deposits under an 8.5% base rate environment, the expected loan portfolio volume is NPR 107.68 Billion.